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Numbers that are the sum of six fifth powers in exactly four ways.
7

%I #6 Jul 31 2021 19:24:17

%S 12047994,20646208,21017489,21300963,21741819,24993485,27669050,

%T 28576064,30193856,30785920,35480456,35735194,36082750,37303264,

%U 39035975,46814942,47963291,50047062,50724345,52987561,53076800,53606848,55101101,56766906,57969327,58125980

%N Numbers that are the sum of six fifth powers in exactly four ways.

%C Differs from A345718 at term 23 because 54827300 = 4^5 + 7^5 + 21^5 + 22^5 + 23^5 + 33^5 = 5^5 + 10^5 + 15^5 + 20^5 + 28^5 + 32^5 = 1^5 + 14^5 + 16^5 + 19^5 + 28^5 + 32^5 = 4^5 + 11^5 + 13^5 + 22^5 + 29^5 + 31^5 = 5^5 + 6^5 + 19^5 + 20^5 + 29^5 + 31^5.

%H Sean A. Irvine, <a href="/A346359/b346359.txt">Table of n, a(n) for n = 1..10000</a>

%e 12047994 is a term because 12047994 = 7^5 + 9^5 + 12^5 + 14^5 + 17^5 + 25^5 = 5^5 + 10^5 + 13^5 + 15^5 + 16^5 + 25^5 = 1^5 + 1^5 + 3^5 + 4^5 + 21^5 + 24^5 = 4^5 + 6^5 + 15^5 + 15^5 + 21^5 + 23^5.

%o (Python)

%o from itertools import combinations_with_replacement as cwr

%o from collections import defaultdict

%o keep = defaultdict(lambda: 0)

%o power_terms = [x**5 for x in range(1, 1000)]

%o for pos in cwr(power_terms, 6):

%o tot = sum(pos)

%o keep[tot] += 1

%o rets = sorted([k for k, v in keep.items() if v == 4])

%o for x in range(len(rets)):

%o print(rets[x])

%Y Cf. A344519, A345718, A345816, A346281, A346358, A346360.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jul 13 2021